Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 9x}{x + 4} = \dfrac{-20}{x + 4}$
Explanation: Multiply both sides by $x + 4$ $ \dfrac{x^2 + 9x}{x + 4} (x + 4) = \dfrac{-20}{x + 4} (x + 4)$ $ x^2 + 9x = -20$ Subtract $-20$ from both sides: $ x^2 + 9x - (-20) = -20 - (-20)$ $ x^2 + 9x + 20 = 0$ Factor the expression: $ (x + 5)(x + 4) = 0$ Therefore $x = -5$ or $x = -4$ At $x = -4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -4$, it is an extraneous solution.